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Magazine Chapter 5: Problem 18
Find the orthogonal projection of v onto the subspace \(W\) spanned by the vectors \(\mathbf{u}_{\mathbf{r}}\) (You may assume that the vectors \(\mathbf{u}_{i}\) are orthogonal. ) $$\mathbf{v}=\left[\begin{array}{r} 3 \\ -2 \\ 4 \\ -3 \end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{l} 1 \\ 1 \\ 0 \\ 0 \end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{r} 1 \\ -1 \\ -1 \\ 1 \end{array}\right], \mathbf{u}_{3}=\left[\begin{array}{l} 0 \\ 0 \\ 1 \\ 1 \end{array}\right]$$
Short Answer
Expert verified
The orthogonal projection of \( \mathbf{v} \) onto \( W \) is \([-0.75, 1.75, 1.75, -0.75]^T\).
Step by step solution
01
Determine Inner Products
Calculate the inner products of \( \mathbf{v} \) with each vector of the basis \( \mathbf{u}_1, \mathbf{u}_2, \) and \( \mathbf{u}_3 \). The inner product for vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by \( \langle \mathbf{a}, \mathbf{b} \rangle = \mathbf{a}^T \mathbf{b} \). \( \langle \mathbf{v}, \mathbf{u}_1 \rangle = 3\cdot1 + (-2)\cdot1 + 4\cdot0 + (-3)\cdot0 = 1 \)\( \langle \mathbf{v}, \mathbf{u}_2 \rangle = 3\cdot1 + (-2)(-1) + 4(-1) + (-3)\cdot1 = -5 \)\( \langle \mathbf{v}, \mathbf{u}_3 \rangle = 3\cdot0 + (-2)\cdot0 + 4\cdot1 + (-3)\cdot1 = 1 \)
02
Compute Projections
Project \( \mathbf{v} \) onto each basis vector \( \mathbf{u}_i \). The projection formula is: \( \text{proj}_{\mathbf{u}_i}(\mathbf{v}) = \frac{\langle \mathbf{v}, \mathbf{u}_i \rangle}{\langle \mathbf{u}_i, \mathbf{u}_i \rangle} \mathbf{u}_i \).\( \text{proj}_{\mathbf{u}_1}(\mathbf{v}) = \frac{1}{1^2 + 1^2} \begin{bmatrix} 1 \ 1 \ 0 \ 0 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 \ 1 \ 0 \ 0 \end{bmatrix} = \begin{bmatrix} 0.5 \ 0.5 \ 0 \ 0 \end{bmatrix} \)\( \text{proj}_{\mathbf{u}_2}(\mathbf{v}) = \frac{-5}{1^2 + (-1)^2 + (-1)^2 + 1^2} \begin{bmatrix} 1 \ -1 \ -1 \ 1 \end{bmatrix} = \frac{-5}{4} \begin{bmatrix} 1 \ -1 \ -1 \ 1 \end{bmatrix} = \begin{bmatrix} -1.25 \ 1.25 \ 1.25 \ -1.25 \end{bmatrix} \)\( \text{proj}_{\mathbf{u}_3}(\mathbf{v}) = \frac{1}{1^2 + 1^2} \begin{bmatrix} 0 \ 0 \ 1 \ 1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 0 \ 0 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0.5 \ 0.5 \end{bmatrix} \)
03
Sum the Projections
The orthogonal projection of \( \mathbf{v} \) onto the subspace \( W \) is the sum of the individual projections onto the basis vectors:\( \mathbf{v}_W = \text{proj}_{\mathbf{u}_1}(\mathbf{v}) + \text{proj}_{\mathbf{u}_2}(\mathbf{v}) + \text{proj}_{\mathbf{u}_3}(\mathbf{v}) \)\[ \mathbf{v}_W = \begin{bmatrix} 0.5 \ 0.5 \ 0 \ 0 \end{bmatrix} + \begin{bmatrix} -1.25 \ 1.25 \ 1.25 \ -1.25 \end{bmatrix} + \begin{bmatrix} 0 \ 0 \ 0.5 \ 0.5 \end{bmatrix} = \begin{bmatrix} -0.75 \ 1.75 \ 1.75 \ -0.75 \end{bmatrix} \]
04
Finalize the Solution
The orthogonal projection of \( \mathbf{v} \) onto the subspace \( W \) is really just computing a vector in \( W \) closest to \( \mathbf{v} \). The result from the sum in the previous step provides this vector.The solution is: \[ \begin{bmatrix} -0.75 \ 1.75 \ 1.75 \ -0.75 \end{bmatrix} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inner Products
In linear algebra, the concept of an inner product is pivotal when dealing with vector spaces. An inner product is a mathematical operation that multiplies corresponding components of two vectors and sums the results. Often, this process is akin to finding the "dot product" but generalized for more complex scenarios. For two vectors, \( \mathbf{a} \) and \( \mathbf{b} \), the inner product is notated as \( \langle \mathbf{a}, \mathbf{b} \rangle \), which can also be represented as the transpose of \( \mathbf{a} \) multiplied by \( \mathbf{b} \): \[ \langle \mathbf{a}, \mathbf{b} \rangle = \mathbf{a}^T \mathbf{b} \] This simple yet powerful operation allows us to measure angles and similarities between vectors, which is crucial in identifying components of one vector aligned along another, or orthogonal components. The inner product is essential in projections to find the shadow or closest point of one vector along or on another vector. It also plays an instrumental role in calculating norms, which measure the length of a vector within its space.
Projection Formula
The projection formula is applied when we need to find the extent to which a vector projects onto another vector or a subspace. The orthogonal projection is the prime way to determine how much of a vector lies in the direction of another vector or within a span of multiple vectors. Given a vector \( \mathbf{v} \) and another vector \( \mathbf{u} \), the orthogonal projection of \( \mathbf{v} \) onto \( \mathbf{u} \) utilizes the formula: \[ \text{proj}_{\mathbf{u}}(\mathbf{v}) = \frac{\langle \mathbf{v}, \mathbf{u} \rangle}{\langle \mathbf{u}, \mathbf{u} \rangle} \mathbf{u} \]
- This formula calculates the component of \( \mathbf{v} \) that aligns with \( \mathbf{u} \).
- The denominator \( \langle \mathbf{u}, \mathbf{u} \rangle \) is often referred to as the "norm squared" of \( \mathbf{u} \), ensuring the projection scales properly.
- The inner product, \( \langle \mathbf{v}, \mathbf{u} \rangle \), identifies how much \( \mathbf{v} \) extends along \( \mathbf{u} \).
Orthogonal Subspaces
In the exploration of vector spaces, orthogonal subspaces are subsets of a vector space that are at right angles to each other. Essentially, vectors from these subspaces have an inner product of zero, indicating no mutual influence in their directions. When vectors are orthogonal, their interactions in terms of projections simplify extensively, as their influences are isolated and summed up separately. Each vector in an orthogonal set contributes uniquely to a subspace without overlap, ensuring the maximum use of dimensions in any calculation. In our exercise, the vectors \( \mathbf{u}_1, \mathbf{u}_2, \text{and} \mathbf{u}_3 \) form an orthogonal basis for the subspace \( W \). This orthogonality ensures that the projection calculations are straightforward and the sum of individual projections gives the complete projection of \( \mathbf{v} \) in the subspace \( W \). The concept of orthogonal subspaces is vital in simplifying problems and ensuring clarity in determining the exact contributions of each vector.
Linear Algebra Problem Solving
The discipline of linear algebra opens up a range of strategies for solving problems related to vectors and matrices. It is a cornerstone in fields such as physics, computer graphics, and machine learning, offering tools and methods for simplifying complex multidimensional problems.
- Approaching problems via vector spaces allows the breakdown into recognizable patterns, such as projections and transformations, ensuring structured answers.
- Linear algebraic techniques like calculating inner products or using projection formulas offer precision in tasks ranging from vector projection to determining vector norms.
- Solving linear algebra problems often involves understanding properties of matrices, eigenvectors, and eigenvalues, enabling robust solutions to real-world problems.
